A rod of length \(10\) cm lies along the principle axis of a concave mirror of focal length \(10\) cm as shown in figure. The length of the image is _____ cm.

For the given concave mirror, the focal length is \(f = -10\) cm.

The rod lies along the principal axis. From the figure, the near end of the rod is at a distance of \(20\) cm from the pole of the mirror. Since the length of the rod is \(10\) cm, the far end is at a distance of \(20 + 10 = 30\) cm from the pole.

Let the near end be \(A\) and the far end be \(B\).
Object distance for end \(A\), \(u_A = -20\) cm.
Object distance for end \(B\), \(u_B = -30\) cm.

Using the mirror formula, \(\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\)

For end \(A\):
\(\dfrac{1}{v_A} + \dfrac{1}{-20} = \dfrac{1}{-10}\)
\(\dfrac{1}{v_A} = \dfrac{1}{20} - \dfrac{1}{10} = -\dfrac{1}{20}\)
\(v_A = -20\) cm

For end \(B\):
\(\dfrac{1}{v_B} + \dfrac{1}{-30} = \dfrac{1}{-10}\)
\(\dfrac{1}{v_B} = \dfrac{1}{30} - \dfrac{1}{10} = \dfrac{1 - 3}{30} = -\dfrac{2}{30} = -\dfrac{1}{15}\)
\(v_B = -15\) cm

The length of the image is the distance between the images of the two ends:
\(\text{Length of image} = |v_A - v_B| = |-20 - (-15)| = |-5| = 5\) cm.

Answer: \(5\)